This is my working 2-column proof for Book 1 Proposition 7. I would be remiss in saying that this is completely foolproof. One question is how we are to formulate a proof by contradiction within the structure of a 2-column proof; needless to say "For contradiction" as a justification on first glance does not look as mathematically salient as such a justification could be. Secondly, I haven't a justification for (7); I would like to keep the justifications as Euclidean as possible, and you can see that it is the transitive property of inequality, but Euclid does not explicitly give this in book 1 despite C.N. 1 being the transitive property of equality. The justification could very well be "Transitive Property of Inequality" but that defeats my purpose of constructing the 2-column proofs using what Euclid has defined, postulated, notioned, and proved previously
\begin{array}{l l} \text{If } \triangle ABC \text{ and } \triangle ABD \text{ stand on the same base } AB \text{ and }\\[4pt] \text{on the same side of it, and if } AC = AD \text{ and } BC = BD, \text{ then } C = D.\\[4pt]\\ \text{Statement} & \text{Justification} \\[6pt] (1) \text{ Let }\triangle ABC\text{ and }\triangle ABD\text{ on base }AB,\ AC=AD,\ BC=BD. & \text{Given.} \\[4pt] (2) \text{ Assume }C\neq D. & \text{For contradiction.} \\[4pt](3) \text{ Draw }CD. & \text{Postulate 1}\\[4pt](4) \angle ACD=\angle ADC. &\text{I.5}\\[4pt] (5)\angle ADC > \angle DCB. &\text{Common Notion 5}\\[4pt] (6)\angle BDC > \angle ADC &\text{Common Notion 5}\\[4pt] (7)\angle BDC \gg\angle DCB \\[4pt] (8) \text{ But }\angle BDC = \angle DCB. &\text{I.5}\\[4pt](9) \therefore C=D. &Q.E.D \end{array}
